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1997. First Day Where You Have Been in All the Rooms

Difficulty Topics

Description

There are n rooms you need to visit, labeled from 0 to n - 1. Each day is labeled, starting from 0. You will go in and visit one room a day.

Initially on day 0, you visit room 0. The order you visit the rooms for the coming days is determined by the following rules and a given 0-indexed array nextVisit of length n:

  • Assuming that on a day, you visit room i,
  • if you have been in room i an odd number of times (including the current visit), on the next day you will visit a room with a lower or equal room number specified by nextVisit[i] where 0 <= nextVisit[i] <= i;
  • if you have been in room i an even number of times (including the current visit), on the next day you will visit room (i + 1) mod n.

Return the label of the first day where you have been in all the rooms. It can be shown that such a day exists. Since the answer may be very large, return it modulo 109 + 7.

 

Example 1:

Input: nextVisit = [0,0]
Output: 2
Explanation:
- On day 0, you visit room 0. The total times you have been in room 0 is 1, which is odd.
  On the next day you will visit room nextVisit[0] = 0
- On day 1, you visit room 0, The total times you have been in room 0 is 2, which is even.
  On the next day you will visit room (0 + 1) mod 2 = 1
- On day 2, you visit room 1. This is the first day where you have been in all the rooms.

Example 2:

Input: nextVisit = [0,0,2]
Output: 6
Explanation:
Your room visiting order for each day is: [0,0,1,0,0,1,2,...].
Day 6 is the first day where you have been in all the rooms.

Example 3:

Input: nextVisit = [0,1,2,0]
Output: 6
Explanation:
Your room visiting order for each day is: [0,0,1,1,2,2,3,...].
Day 6 is the first day where you have been in all the rooms.

 

Constraints:

  • n == nextVisit.length
  • 2 <= n <= 105
  • 0 <= nextVisit[i] <= i

Solution

first-day-where-you-have-been-in-all-the-rooms.py
class Solution:
    def firstDayBeenInAllRooms(self, nextVisit: List[int]) -> int:
        n = len(nextVisit)
        dp = [0] * n

        for i in range(1, n):
            dp[i] = (2 * dp[i - 1] - dp[nextVisit[i - 1]] + 2) % (10 ** 9 + 7)

        return dp[-1]